Answer:
0.1127 grams of Mg is used to produce one gram of silver
Explanation:
Write a balanced equation
Mg(s) +2Ag + (aq) --> 
(aq) + 2Ag(s)
The atomic mass of magnesium is 24.31 g/mol and the atomic mass of silver is 107.84g/mol
1 g Ag of silver is produced from [(1 mol Mg)(24.31 g/mol)] /[ (2mol Ag)(107.84g/mol)]
1 g Ag of silver is produced from 0.1127 grams of Mg
0.1127 grams of Mg is used to produce one gram of silver
Answer:
The 
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Explanation:
The expression of the equilibrium constant of base 
can be given as:
![K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%5BH_2O%5D%7D)
]![K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%20%5BH_2O%5D%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
As we know, water is pure solvent, we can put ![[H_2O]=1](https://tex.z-dn.net/?f=%5BH_2O%5D%3D1)
![K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%201%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
So, the the expression for the weak base equilibrium is:
Answer:
Rutherford's theory laid the foundation upon which Bohr's model is founded. Rutherford established the fact that at the center of the atom, there is a nucleus whose radius is smaller than the radius of the atom. This nucleus is positively charged and most of the mass of the atom is concentrated there. Electrons move round this nucleus in orbits.
The experimental evidences of the Bohr's model shows that Rutherford's model was fundamentally correct. However, Bohr's model introduced the idea of quantization of the energy of electrons in an atom. The model went further to explain the spectra lines of the hydrogen atom.
Explanation:
57 degrees celcius is equal to 330 degrees kelvin
67.88 kPa is equal to 67880 Pa
85.3 liters is equal to 0.0853 m^3
Now, the equation we will use to solve this question is:
PV = nRT where:
P is the pressure of gas = 67880 Pa
V is the volume of gas = 0.0853 m^3
n is the number of moles we are looking for
R is the gas constant = 8.31441 J K-1<span> mol</span><span>-1
T is the temperature of gas = 330 degrees kelvin
Substitute with the givens in the above equation to get n as follows:
n = (PV) / (RT)
n = (67880*0.0853) / (8.31441*330)
n = 2.11 moles</span>
